Answer to Question #148130 in Discrete Mathematics for Promise Omiponle

Question #148130
Let A be a countable set, and B is another set. Assume further that there exists an onto function f:A->B. Is B necessarily countable? Provide a full justification for your answer.
1
Expert's answer
2020-12-18T14:47:45-0500

How the proof proceeds depends on how you define "countable". A nonempty set "B"

is countable if either:

1) there exists a one-to-one function "f:B\\to \\mathbb{N}"

or

2) there exists an onto function "g:\\mathbb{N} \\to B"

It turns out these are equivalent, so go with the second.

Since "A" is countable there exists an onto function "g:\\mathbb{N}\\to A" , and by hypothesis there is an onto function "f:A\\to B" . The composition "f \u2218g:\\mathbb{N}\\to B"  is onto, so that B

B is countable.


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Comments

Assignment Expert
01.12.20, 20:13

Dear Promise Omiponle, your questions will be evaluated by our experts. We shall try to solve and publish the answer as soon as possible, but we cannot assure the exact time frame for a solution of the question. If you have several urgent questions and you need a solution with the strict deadline, you also can submit an order as well.

Promise Omiponle
01.12.20, 08:04

Hello. I just wanted to know if it would be possible for at least half of the 28 questions I uploaded, including this one, to be answered before Thursday, December 3rd 2020? I know this may seem like a lot to ask and you may suggest "make a full order request" but that's not an option for me sadly.

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